kinematics  
 
  Eddy currents  
 
  Lenz's law  
 
  Lines of fields  
 
  Insulators  
 
  Diff. Eqs   
 
  Constants  
 
  Units   
 
  home  
 
  ask us  
 

 








© The scientific sentence. 2010

Ampere's law




1. Ampere's law: Circular path



We have seen that the magnetic field produced by a long straight wire crossed by a current I at a point P in the axis of the loop is:

B = (μoI/2πR)

R is the perpendicular distance from the wire to the point P. The direction of the magnetic field B is tangent to a line of magnetic field. Each line closes on itself and encircles the current carrying wire. The related circle is in a plane perpendicular to the wire. This circle is considered as a closed path, and its circumference 2πR is equal to ∫dr; where dr is the distant element along the circumference of radius R. Hence the above relation becomes:

B 2πR = B ∫dr = ∫Bdr μoI

Using a dot product, we generalize this result to a line integral of B formula called Ampere's law:

∫ B.dr μoI
dr is the infinitesimal displacement along the closed path. The line integral does not depend on the radius of the closed circle.

     →    
   B ⋅  dr   =   μo I  



2. Ampere's law: Arbitrary closed path



Consider another closed path linked by the current in a long straight wire. The closed path lies in the plane perpendicular to the axis of the wire. It consists of several circular arcs centered on the axis of the wire and several radial lines that connect the arcs.

Since the line integral of B does not depend on the distance (radius) of the circular arcs, and the dot product of B and the radial lines is zero, the contributions to the magnetic field around the wire is the sun (integral) along the circular arcs ln(arc length subtended the angle θn = ln/Rn):

∮Bdr = ∮(1)B1dr + ∮(2)B2dr + ... ∮(n)Bn dr =
B1∮(1)dr + B2∮(2)dr + ... Bn∮(n)dr =
l1B1 + l2B2 + ... + lnBn = ΣlnBn

We have:

circular arc n = ln = Rnθn, and
Bn = μoI/2πRn

Hence

lnBn = Rnθn μoI/2πRn = θn μoI/2π

Σ lnBn = μoI/2π Σ θn = μoI/2π (2π) = μoI

Therefore:

     →    
   B. dr   =   μoI
           

3. Ampere's law: Path not linked by current




Now, we consider a closed path that is not linked by the current in the long straight wire. In the case (a) of the closed path with two circular arcs that subtend the same angle θ at their common center, and two radial lines, the contribution of the radial lines is again zero, and the only contribution comes from the circular arcs part:

∮B.dr = ∮B.dr + 0 + ∮B.dr + 0 =
B1 l1 - B2 l2 = μoI/2πRn

We have:

l1 = R1θ,
l2 = R2θ

and
B1 = μoI/2πR1
B2 = μoI/2πR2

B1 l1 = μoI l1 /2πR1 =
μoI l1 l2 /2πR2l1 =
μoI l2 /2πR2 = B2 l2

Therefore

∮B.dr = 0



For the path that is not linked by the current: the line integral of B around any closed path is zero:
 →    →
   ∮ B . dr = 0



4. General expression of Ampere's law




Consider an arbitrary closed path near currents of conductors of any form. The arbitrary closed path is linked by some currents (I1, I2, I3, I4) and not linked by some others (I5, I6). The sum Σi of the currents that link the closed path gives the general line integral of the magnetic field B around this path:

     →    
   B. dr   =   μoΣi  
           



5. Ampere's law and the displacement current

Let' rewrite Ampere's law:

 →    →
      ∮B . dr = μoΣi



Note that the origin of this formula comes from the result of considering a long straight wire. The question is what about the fact that the wire is not long? Secondly, this law involves curved path that is linked by the current piercing a surface without specifying that surface.
The inconsistency of this law was raised by James Clerk Maxwell in 1861 because he needed something else to complete his equations for the electromagnetic waves. He suggested adding another current called displacement current Id to the conduction current I in order to write a new law called Maxwell-Ampere law . The expression of Id is:

Id = εoE/dt

ΦE is the flux of the electric field E through the surface bounded by the closed path linked by the displacement current Id.

The only given proof to bring out the existence of this current is to consider a capacitor. If a capacitor is being charged, then there is a change in the electric field E between the plates of the capacitor due to the accumulation of charges Q on the plates of area S = Q/σ where σ is the surface charge density.

Since E = σ/εo = Q/Sεo, the displacement current is Id = dQ/dt = d(ESεo)/dt = εo d(ES)/dt = εoE/dt.

     →    
   B ⋅  dr   =   μo (I + εoE/dt)  
 






 


chimie labs
|
Physics and Measurements
|
Probability & Statistics
|
Combinatorics - Probability
|
Chimie
|
Optics
|
contact
|


© Scientificsentence 2010. All rights reserved.